Difference between revisions of "2019 AMC 8 Problems/Problem 22"
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==Solution 2 (Answer options)== | ==Solution 2 (Answer options)== | ||
We can try out every option and see which one works out. By this method, we get <math>\boxed{\textbf{(E)}\ 40}</math>. | We can try out every option and see which one works out. By this method, we get <math>\boxed{\textbf{(E)}\ 40}</math>. | ||
Revision as of 19:59, 15 January 2022
Contents
Solution 2 (Answer options)
We can try out every option and see which one works out. By this method, we get 
.
Solution 3
Let x be the discount. We can also work in reverse such as (
)
= 
.
Thus 
= 
. Solving for 
gives us 
. But has to be positive. Thus = 
.
Solution 4 ~ using the answer choices
Let our original cost be 
We are looking for a result of 
then. We try 16% and see it gets us higher than 84. We try 20% and see it gets us lower than 16 but still higher than 84. We know that the higher the percent, the less the value. We try 36, as we are not progressing much, and we are close! We try 
, and we have the answer; it worked.
Video explaining solution
https://www.youtube.com/watch?v=_TheVi-6LWE - Happytwin
Associated video - https://www.youtube.com/watch?v=aJX27Cxvwlc
https://www.youtube.com/watch?v=RcBDdB35Whk&list=PLLCzevlMcsWNBsdpItBT4r7Pa8cZb6Viu&index=4 ~ MathEx
~savannahsolver
https://www.youtube.com/watch?v=DaF8uD8V8u0&list=PLbhMrFqoXXwmwbk2CWeYOYPRbGtmdPUhL&index=24
See Also
| 2019 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 21 |
Followed by Problem 23 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
| All AJHSME/AMC 8 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
